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2 years ago

CAN SOMEONE PLEASE HELP ME RIGHT NOW.

Mathematics

1 answer:

exis [7]2 years ago

6 0

Answer:

m∠SRV = 48°

Step-by-step explanation:

In the parallelogram attached,

m∠TUV = 78°

m∠TVU = 54°

By applying the property of the angles of a triangle in ΔTVU,

m∠TUV + m∠TVU + m∠UTV = 180°

78° + 54° + m∠UTV = 180°

m∠UTV = 180° - 132°

= 48°

Sides RS and TU are the parallel sides of the parallelogram and diagonal TR is a transverse.

Therefore, ∠UTV ≅ ∠SRV [Alternate interior angles]

m∠UTV = m∠SRV = 48°

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Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect

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Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

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$\bf (cos(\pi t), sin(\pi t), t)$

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$\bf cos^2(\pi t)+sin^2(\pi t)=1$

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The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface

$\bf z=x^2+y^2$

at the point (-1, 0, 1) is given by

$\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)$

where

$\bf f(x,y)=x^2+y^2$

since

$\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y$

so, a normal vector to the tangent plane is

(-2,0,-1)

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$\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)$

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.

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$\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta$

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$\bf \theta$ = angle between the tangent vector and its projection onto the tangent plane. So

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and

$\bf \theta=arccos(0.8038)=0.6371\;radians$

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